In an experiment, 4g of `M_(2)O_(x)` oxide was reduced to 2.8g of the metal. If the atomic mass of the metal is `56g"mol"^(-1)`, the number of oxygen atoms in the oxide is:

Options

A. 1
B. 2
C. 3
D. 4

Answer

C

Explanation

1 mole of `M_(2)O_(x)` on reduction gives 2 g atoms of M
i.e. `(2xx56+16x)g M_(2)O_(x)" gives M"=2xx56g`
`= 112g`
`therefore 4g M_(2)O_(3)" will give"=(112)/(112+16x)xx6g`
`therefore" "(112xx4)/(112+16x)=28`
i.e., `448=2.87(112+16x)`
or `112+16x=160" or "x=3.`